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Binomial theorem to evaluate limits?

  • #1

Homework Statement



lim x->1 (X^9 + x -2)/(x^4 + x -2)

I know how to do this using L'Hopitals Rule and I get 2

Homework Equations



(1+b)^n = 1 + bn + n(n-1)b^2/2! + n(n-1)(n-2)b^3/3! ....

The Attempt at a Solution



Let x = h+1

x -> 1
h -> 0

lim h->0 (h+1)^9 + h-1/(h+1)^4+h-1

lim h->0 h^9 (1+1/h)^9 +h-1/h^4(1+1/h)^4 +h-1

Binomial theorem:

(1+1/h)^9 = 1 +9/h +36/h^2 + 84/h^3 ...

(1+1/h)^4 = 1 +4/h + 6/h^2 + 4/h^3 ...

lim h->0 h^9(1 +9/h +36/h^2 + 84/h^3 ...) +h-1/h^4(1 +4/h + 6/h^2 + 4/h^3 ...) +h-1


This is stil ∞.0 right?

I tried to get rid of the +h-1 at the end by doing

h^7(h^2 +10h +35 +84/h...)/h^2(h^2+5h+5+4/h...)

but then you get ∞.∞
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,833
955
You are fine to
lim h->0 ((h+1)^9 + h-1)/((h+1)^4+h-1)
But I don't understand why you then switched to "1/h". Taking h to 0 is much simpler than taking 1/h to infinity. By the binomial theorem, (h+ 1)^9= h^9+ 9h^8+ ...+ 36h^2+ 9h+ 1 so that (h+1)^9+ h- 1= h^9+ 9h^8+ ...+ 36h^2+ 10h. Similarly (h+ 1)^4= h^4+ 4h^3+ 6h^2+ 4h+ 1 so that (h+ 1)^4+ h- 1= h^4+ 4h^3+ 6h^2+ 5h.

The fraction can be written
[tex]\frac{h^9+ 9h^9+ ...+ 36h^2+ 10h}{h^4+ 4h^3+ 6h^2+5h}= \frac{h(h^8+ 9h^7+ ...+ 36h+ 10)}{h(h^3+ 4h^2+ 6h+ 5)}[/tex]
Now those first "h" terms cancel and you can just take h= 0.

Are you required to use that method? Simpler than that or L'Hopital:

The only problem here is that both numerator and denominator are 0 at x= 1. But since they are polynomials, that means that each has x- 1 as a factor:
[tex]x^9+ x- 2= (x- 1)(x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2)[/tex]
and
[tex]x^4+ x- 2= (x- 1)(x^3+ x^2+ x+ 2)[/tex]
So you just need to evaluate
[tex]\frac{x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2}{x^3+ x^2+ x+ 2}[/tex]
at x= 1.
 
  • #3
You are fine to
lim h->0 ((h+1)^9 + h-1)/((h+1)^4+h-1)
But I don't understand why you then switched to "1/h". Taking h to 0 is much simpler than taking 1/h to infinity. By the binomial theorem, (h+ 1)^9= h^9+ 9h^8+ ...+ 36h^2+ 9h+ 1 so that (h+1)^9+ h- 1= h^9+ 9h^8+ ...+ 36h^2+ 10h. Similarly (h+ 1)^4= h^4+ 4h^3+ 6h^2+ 4h+ 1 so that (h+ 1)^4+ h- 1= h^4+ 4h^3+ 6h^2+ 5h.

The fraction can be written
[tex]\frac{h^9+ 9h^9+ ...+ 36h^2+ 10h}{h^4+ 4h^3+ 6h^2+5h}= \frac{h(h^8+ 9h^7+ ...+ 36h+ 10)}{h(h^3+ 4h^2+ 6h+ 5)}[/tex]
Now those first "h" terms cancel and you can just take h= 0.

Are you required to use that method? Simpler than that or L'Hopital:

The only problem here is that both numerator and denominator are 0 at x= 1. But since they are polynomials, that means that each has x- 1 as a factor:
[tex]x^9+ x- 2= (x- 1)(x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2)[/tex]
and
[tex]x^4+ x- 2= (x- 1)(x^3+ x^2+ x+ 2)[/tex]
So you just need to evaluate
[tex]\frac{x^8+ x^7+ x^6+ x^5+ x^4+ x^3+ x^2+ x+ 2}{x^3+ x^2+ x+ 2}[/tex]
at x= 1.
How did you divide x^4 +x -2 by (1-x)? When I did it I got (-x^3+x^2+x) -2/(1-x)
I did the polynomial division method.

Also, thank you.
 
  • #4
How did you divide x^4 +x -2 by (1-x)? When I did it I got (-x^3+x^2+x) -2/(1-x)
I did the polynomial division method.

Also, thank you.
Sorry, I mean how did you divide x^4 +x -2 by (x-1)? I got (x^3 -x^2 -x -2/(x-1) )

and for x^9 +x -2 divided by (x-1) I got (x^8-x^7 -x^6 -x^5 -x^4 -x^3 -x^2 -x -2/(x-1))
 
  • #5
epenguin
Homework Helper
Gold Member
3,713
753
Sorry, I mean how did you divide x^4 +x -2 by (x-1)? I got (x^3 -x^2 -x -2/(x-1) )

and for x^9 +x -2 divided by (x-1) I got (x^8-x^7 -x^6 -x^5 -x^4 -x^3 -x^2 -x -2/(x-1))
Whether they are true is something you can more easily check than how it was got.
 

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